The convolution integral when— f1(t) = e

We know that
f₁(t)*f₂(t) = ∫^t₀ f₁(τ ).f₂(t – τ )dτ
= ∫^t₀ 2τ·e^-2(t – τ) dτ
= e^{- 2t} ∫^t₀ 2τ·e^2τdτ

= 2e^{- 2t}		τ·	e^2τ	∫^t₀ τ·	e^2τ	· dτ		^t
			2		2			₀

= 2e^{- 2t}			τe^2τ	–	e^2τ			^t
			2		4			₀

= 2e^{- 2t}		te²	–	e^2t	+	1
		2		4		4

=		t	1	+	e^{- 2t}			u(t)
			2		2

Hence, alternative (A) is the correct answer.